Add graph to Serway and Jewett v8's 25.27.b solution.

diff --git a/latex/problems/Serway_and_Jewett_8/problem25.27.tex b/latex/problems/Serway_and_Jewett_8/problem25.27.tex
index e4697f647450b4873fa68b19de08966f2d9f6b42..31b05f609286ea762c79ee235d9fe50a7e50b920 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem25.27.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem25.27.tex
@@ -49,8 +49,45 @@ Conserving energy
 \end{align}
 
 \Part{b}
-The speeds for conductors would be \ans{greater}, because as the
-spheres close the charge polarization will intensify, leading to a
-greater effective $\Delta r_{12}$ over a more sensitive portion of the
-$1/r$ voltage curve.
+The speeds for conductors would be \ans{greater}, because the opposite
+charges on each sphere would be drawn to the inner walls, reducing the
+effective distance between the attracting charges and increasing the
+force.
+
+Thinking about the problem in terms of electric potential, the
+difference between the initial and final effective distances would be
+greater ($\Delta r_{12}' = |r_{12,f}' - r_{12,i}'| > \Delta r_{12}$)
+because as the spheres close the charge polarization will intensify.
+This alone would increase the electric potential converted to kinetic
+energy, but the distance change also occurs over a more sensitive
+portion of the $1/r$ voltage curve.
+\begin{center}
+\begin{asy}
+import graph;
+
+size(6cm, 4cm, IgnoreAspect);
+
+real r_min = 0.5;
+real r_max = 2;
+real r12i = 1.6;
+real r12f = 0.9;
+real r12ip = 1.4;
+real r12fp = 0.6;
+
+real V(real r)
+{
+  return 1/r;
+}
+
+draw((r12i, V(r12i))--(r12f, V(r12i))--(r12f, V(r12f)), blue);
+draw((r12ip, V(r12ip))--(r12fp, V(r12ip))--(r12fp, V(r12fp)), green);
+draw(graph(V, r_min, r_max), red);
+label("$\Delta r_{12}$", ((r12i+r12f)/2, V(r12i)), align=S, p=blue);
+label("$\Delta r_{12}'$", ((r12ip+r12fp)/2, V(r12ip)), align=SW, p=green);
+label("$\Delta V_{12}$", (r12f, (V(r12i)+V(r12f))/2), align=W, p=blue);
+label("$\Delta V_{12}'$", (r12fp, (V(r12ip)+V(r12fp))/2), align=W, p=green);
+xaxis("$r$", xmin=0, xmax=r_max, LeftTicks(NoZero));
+yaxis("$\frac{V}{kq}$", ymin=0, ymax=V(r_min), RightTicks);
+\end{asy}
+\end{center}
 \end{solution}